Question

How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?

HCl(aq)+NaOH(aq)⟶NaCl(aq)+H2O(l)ΔH°=−58kJ

a) 145 kJ
b) 58 kJ
c) -58 kJ
d) -145 kJ

Answer 1

When 100 mL of 0.250 M HCl and 200 mL of 0.150 M NaOH are mixed, the heat produced can be calculated using the equation q = m ÐTxC. The correct answer is -58 kJ.

Step-by-step explanation:

To calculate the amount of heat produced when 100 mL of 0.250 M HCl and 200 mL of 0.150 M NaOH are mixed, we can use the equation q = m ÐTxC. First, we need to calculate the mass of each solution by multiplying the volume by the density. For HCl: 100 mL * 1.00 g/mL = 100 g. And for NaOH: 200 mL * 1.00 g/mL = 200 g. Next, we can calculate the change in temperature (ÐT) by subtracting the initial temperature from the final temperature: ÐT = 28.9 °C - 22.0 °C = 6.9 °C.

Finally, we can calculate the heat using the specific heat capacity of water (4.18 J/g°C): q = (100 g + 200 g) * 6.9 °C * 4.18 J/g°C = 9636 J. However, since the equation for the reaction given has a heat change of -58 kJ per mole of HCl, we need to convert the heat from joules to kilojoules and adjust for the moles of HCl: q = 9636 J / 1000 J/kJ * (0.1 mol HCl / 1 mol) = -0.9636 kJ. Therefore, the correct answer is option c) -58 kJ.